0
$\begingroup$

I am analyzing a signal with FFT and I found an effect, that I do not understand. I am expecting to have an amplitude modulated signal. As far as I know, if you compute the power of the FFT of an amplitude-modulated signal, you observe sidebands to the left and right of the carrier frequency.

E.g., if you have a sinusoidal carrier at frequency $w_c$, and you modulate the carrier's amplitude with frequency $w_m$, you see sidebands appearing at $w_c \pm w_m$.

However, in my signal's power spectrum, I see the following. Note, that the x-axis shows not the absolute frequency, but the delta from the carrier.

Power spectrum of an amplitude-modulated signal

At the frequency, where I would expect peaks in the spectrum, I see drops. How can that be? What does that tell me about the signal? If I look at the signal in the time domain. I can see, that amplitude modulation is actually happening!

Improve this question
$\endgroup$
4
  • $\begingroup$ Why is your axis "frequency deviation"? To do what you describe in your question, you need to plot using plain frequency. $\endgroup$
    – MBaz
    Commented Mar 13, 2020 at 13:45
  • $\begingroup$ Can you post a snippet of your code? $\endgroup$
    – DSP Novice
    Commented Mar 13, 2020 at 13:50
  • $\begingroup$ Can you tell us what signal you're amplitude modulating, and maybe show it to us in the time-domain? If it's supposed to be an AM-modulated sine wave, something is definitely wrong. $\endgroup$
    – TimWescott
    Commented Mar 13, 2020 at 13:54
  • $\begingroup$ Is this a real world signal or simulated in MATLAB? Reason I am asking is, when you generate a real world carrier signal cos(omega_c), it can start the Local Oscillator with random phase. So x_c = cos(omega_c + phi). It may happen that phi=180 degrees. So your x_c and its modulating signal will be out of phase (cos (theta+pi) = -cos(theta) ) which will negatively affect the total amplitude. $\endgroup$
    – jithin
    Commented Mar 13, 2020 at 17:34

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.